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Spring 2007 - Linshaw's Class - Quiz 4

Spring 2007 - Linshaw's Class - Quiz 4 - 2 − 3 = − 26...

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Name: PID: TA: Sec. No: Sec. Time: Math 20C. Quiz 4 May 18, 2007 1. Using Lagrange multipliers, fnd the maximum and minimum values oF the Function f ( x, y ) = 4 x + 6 y on the circle x 2 + y 2 = 13. Solution: Let g ( x, y ) = x 2 + y 2 . Setting f ( x, y ) = λ g ( x, y ), we obtain a 4 , 6 A = λ a 2 x, 2 y A . So we need to solve the system oF equations 4 = 2 λx, 6 = 2 λy, x 2 + y 2 = 13 . Note that λ n = 0 since 4 = 2 λx n = 0, so we can divide by λ in the frst two equations. We obtain x = 2 λ and y = 3 λ , which we then substitute in the equation x 2 + y 2 = 13. We get 4 λ 2 + 3 λ 2 = 13, so λ 2 = 1. Hence λ = 1 or λ = 1. IF λ = 1, we get x = 2 and y = 3. IF λ = 1, we get x = 2 and y = 3. ±inally, f (2 , 3) = 26 and f (
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Unformatted text preview: 2 , − 3) = − 26. So the maximum value oF f on the circle is 26 and the minimum is − 26. 2. Evaluate i 2 i 1 (2 x + y ) 4 + 3 y 2 dx dy. Solution: We have i 2 i 1 (2 x + y ) 4 + 3 y 2 dx dy = i 2 A ( y ) dy, where A ( y ) = I 1 (2 x + y ) 4 + 3 y 2 dx . A ( y ) = p (2 x + y ) 5 10 + 3 y 2 x Pv v v v x =1 x =0 = ( y + 2) 5 10 + 3 y 2 − y 5 10 . i 2 ( y + 2) 5 10 + 3 y 2 − y 5 10 = p ( y + 2) 6 60 + y 3 − y 6 60 Pv v v v y =2 y =0 = 4 6 60 + 8 − 2 6 60 − 2 6 60 ....
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